http://quark.physics.uwo.ca/~harwood/public_html/phys2/torque_sample.

doc

March 29, 2005

Solutions are/were in hidden text.

TORQUE SAMPLE PROBLEMS 1. A bike pedal is 45 cm from the axle. How much torque does a 90 lb girl standing on the pedal exert when (i) it is horizontal? (ii) it is at its lowest point? Solution: (i) = rF 90 lb = 90 2.2 kg = 40.9 kg

= 40.9 kg * 9.8 N/kg F = 401 N

= 0.45m 401 N = 180 Nm (ii) = 0. The force is in the same direction as r. no twisting motion. 2. Find the unknown mass needed to balance the mobile.

5cm

20 cm

40 cm

50 cm

5 cm

m1 = ?

mblock= 150g

m2 = 400g

m3 = 100g

Solution: Block is 120 cm long. Mid point is 60 cm from one end. Midpoint is 35 cm to the right of the supporting string. = 0. 1 + bl + 2 + 3 = 0 Since all of the forces are masses multiplied by g and we dont need to find a number for the total torque (we assume that it is zero), we can divide out the g. .: m1r1 + mblrbl + m2r2 + m3r3 = 0 m1(-20cm) + (150g)(+35cm) + (400g)(+40 cm) + (100g)(+90 cm) = 0 m1 = 1510 g

3. A 400 kg beam must be secured as in the diagram. A cable that can hold 1000N before snapping is tied to one end of the 4 m beam. The cable makes a 50 angle with the beam. (a) how much mass must be placed at location A to just balance the beam? A is 0.1 m from the left end of the beam. (b) what is the maximum amount of mass that can be placed at A before the beam tips in the other direction?0.5m 0.5m

Solution: (a) The beam wants to pivot about point C. There are three torques: from mass at A, from rope, and from the mass of the beam. Find distances from each force to point C: rA = 0.9 m rrope = 1.0 m rbeam = +1.0 m (the beam is 4 m long, c.m. is at 2m this is 1.0 m from point C.) Find forces: FA = ? Frope = 1000N sin500 Fbeam = mbeam * g = 766 N = 400kg 9.8N/kg = 3920N Since it is balanced = 0.

A + rope + beam = 0 rAFgA + rropeFrope + rbeamFbeam = 0

(-0.9m)FgA + (-1.0m)(766N) + (1.0m)(3920N) = 0 FA = 3504 N FA = mAg mA = 358 kg <this does seem quite large! check answer> b) The beam will now pivot about point B. Are there two torques or three? Does the rope still pull downwards when the beam is about to tilt? Assume yes. The forces are the same as in part (a) (mass A is unknown), but the distances have changed. we will find the 3 distances again: rA = 0.4 m rrope = 0.5 m rbeam = +1.5 m Since it is balanced = 0. A + rope + beam = 0

rAFgA + rropeFrope + rbeamFbeam = 0

(-0.4m)FA + (-0.5m)(766N) + (1.5m)(3920N) = 0

FA = 13742 N .: mA = 1402 kg. This means that any mass between 358kg and 1402 kg will balance the beam and stop it from tipping over. 4. Martindale p220 # 6 <I havent had time to type in this problem and its solution yet.>